Richard Habermanrhaberma<opost.smu.edu

Chapters 1-5. Chapter 11 on Green's functions for the heat and \vave equation isan exception since it requires Chapters 9 and 10.

New to this edition is Chapter 14 which primarily discusses dispersive waves.Chapter 14 is more advanced, and, unlike the rest of the text, has been written veryconcisel:\'. Nonetheless, it is mostly self-contained and hence accessible to strongundergraduates. Topics discussed include group velocity and envelope equationsfor linear diHpersive waves and solitary waves and solitons for nonlinear dispersivewaves. In addition, instability and bifurcation phenomena for partial differentialequations are discussed as well as perturbation methods (multiple scale and bound­ary layer problems). In Chapter 14, the author has attempted to show the vitalityof the contemporary study of partial differential equations in the context of physicalproblem~.

The author has made an effort to preserve the second edition so that previoususers will find little disruption. The chapter on numerical methods is now Chap­ter 6 and all subsequent chapters have been moved back one. Nearly all exercisesfrom the previolls edition have been retained with no change in the order to facili­tate a transitioll for previous users. There are over 200 figures to illustrate variousconcepts, which were prepared for this edition by the author using MATLAB. Mod­ern technolog:y is especially important in its graphical abilities, and I have tried toindicate throughout the text places \vhere three-dimensional visualization is helpful.

Overall my object has been to explain clearly many fundamental aspects ofpartial differential equations a.s an introduction to this vast and important field.After achieving a certain degree of competence and understanding, the student canuse this text as a reference, but for additional information should be prepared torefer to other books such as the ones cited in the bibliography.

Finally, it is hoped t.hat. this text enables the reader to find enjoyment in thestudy of the relationships between mathematics and the physical sciences.

The author gratefully acknowledges the contributions of the following review­ers of the manuscript: Charles R. MacCluer, Michigan State University; CatherineA. Roberts, Northern Arizona Universi(y; \iVilliarn W. Roberts, Jr., University ofVirginia; Surendra Singh, University of Tulsa; Alexander P. Stone, University ofNew Mexico; Ron Lipsman, University of Maryland at College Park; Robert W.Kolkka, Michigan Technological University; Dahlard L. Lukes, University of Vir­ginia; Jonathan Dimock, SUNY at Buffalo; Donald Schwendeman, Rensselaer Poly­technic Institute; William Ames, University of Waterloo; Robert Seeley, Universityof Massachusettes at Boston; and Arnold L. Villone, San Diego State University.

vVe wish to discuHs the solution of elementar~' problems involving partial differ­entia.I equations, the kinds of problems that arise in various fields of science andengineering. A partial differential equation (PDE) is a mathematical equationcontaining partial derivatives 7 for example,

1

2. Solution

3. Interpretation

We begin by formulating the equations of heat flow describing the transfer ofthermal energy. Heat energy is caused by the agitation of molecular matter. Two

Chapter 1

Heat Equation

au auat + 3 ar = O. (1.1.1)

We could begin our study by determining what. functions u (x, t) sat.isfy (1.1.1).However, we prefer to start by investigating a physical problcrIl. We do this fortwo reasons. First, our mathematical techniques probably will be of greater inter­est to you when it becomes dear that. the.'5e methods analyze physical problem.,.Second, we \vill actually find that physical considerations will motivate many of ourmathematical developments.

Many diverse subject area.5 in engineering and the physieal sciences are domi­nated by the study of partial differential equations. No list could be all-inclusive.However, the following exa,mples should give you a feeling for the type of areasthat are highly dependent on the study of partial differential equations: acoustic6: 1

aerodynamics, elasticity, electrodynamics, fluid dynamiC's, geophysics (seismic wavepropagation), heat transfer, meteorology, oceanography, optics, petroleum engineer­ing, plasma physics (ionized liquids and gases), quantum mechanics.

We will follow a certain philosophy of applied mathematics in which the analysisof a problem will have three stages:

1. Formulation

1.1 Introduction

PrefaceXIV

2 Chapter 1. Reat Equatioll 1.2. Conduction of HeaL in One-Dimension 3

since diP volume of a .slice is A .6..x.

heat energ:v = e(J;, t)A ~:r,

~:£ is excpedillgl~'{ small, then el:r, t) may be approximat.t'd a5 a. c()n~tant thnmghuutt.he volume so that

Heat flux. Thermal energ;y flows to the right or left in a one - dimensionalrod. V\-'e introduce the heat flux

heat energy geller~illside per unit. tir:. J+

heat energy flowingacross boundariesper unit time

heat flux (the amount of thermal energy pel' unitt'ime flowing to the right per unit surface un:;a).

¢(;d)

rate of changeof heat energ,}'in Limp

Thb is called conservation of heat energy. For the small slice. the rate of changeof heat f'llergy is

a .at [e(x. t)A fl.,! ,

where the partial derivative J)t, is llsed hecause x b being held fix(~d.

Conservation of heat energy. The heat energy belwem .r awl ;1' + fl.;rchanges in time du(~ onl.y to heat energy flowing a.croBs the edges (::r and :r +~x) andheat. energy generated in::-;ide (duE' to positive or negative sources of heat, ellt,'rgy).No heat energ)T changefoi are due to flow across the lateral surface, since we }W,\'\'

a.'>::;umf'd t.hat the lateral surface b insulaterl. Tlw fundamental heat fim\' process is<1e~c:rihed by the word cquat,ion

fix, t) "" thermal energy den'itY.l

basic proce~ses take place in order for thermal energy t.o move: cOllductioll and con­vection. Conduction ref-mUs from the collisions of neighboring molc('ulc~ in whichthe kinetic energy of vibration of Olle molecule is transferred to its nearest. neighbor.Thermal energ,y is thus spread by conduction even if the lIlolecule:~; thelnselvE':-; donot move their location appreciably. In addition, if a vibrating molecule move.s fromone region to another. it takeH its thermal energy with it.. This type of movementof thermal enE'rgy b taIled convection. In order to begin our study with relatively~implp problems, \\'C will stud.'" heat flow only in cases in which the conduction ofheat energy is much morc significant than its convection. \Ve ,vill tllU~ think ofheat flow primaril.y in the case of solids, although heat. transfer in fluids (liquidsand ga::-ps) is also primarily by conduction if the fluid volocity is sufficiently sInal I.

1.2 Derivation of the Conduction of Heatin a One-Dimensional Rod

vVe &,;[";urne that all thermal quantities (\re constant acrOl-iS a section; the rod is one­dimensional. The simplel-it way thif-i may be accomplished is to insulate perfectlythe lateral surface area of t,he rod. Then no thermal energy can P<'\."lS through thelateral surface. Thf' dependence on x and t corresponds to H. situation in whichthe rod is not uniformly heated; the thermal energy density varies from one crosssection to another.

Thermal energy density. We lwgin by con,idering a rod of constant cr(),,~

sectional an'a A, oriented in the x-direction (from:£ = 0 to :r = L) as illustratedin Fig. 1.2.1. \V(, temporarily' introduce the amount of thf'rmal energy per ullitvolume as an unknown variable and call it the thermal energy density:

Figure 1.2.1: OllP- dimensional rod \\iith heat energy flowing into and out of a thinslice.

¢(x + ~x.t)A

x=Q

¢(x,t)

x x +.6.x :r=L

If ¢(x, t) < 0, it means that heat energy is flowing to t.he left.. Heat energy flowingper unit time across the boundaries of the slice is ¢(x, t)A - ,p(x + fl.x, t)A, since theheat flux is the flow per unit. surface area and it IIlust be multiplied by' the surface'area. If ¢(x. t) > 0 and ¢(x + fl.:r, t) > 0, as illustrated in Fig. 1.2.1. t.hen t.he heat.energy flowing per unit time at x contributes to an iIH~rease of the heat. energy inthe slice, whereas the heat flow at x + ll~' decreases the heat energy.

Heat sources. We also allow for internal sources of thermal energy:

Heat energy. We cOllsider a thin slice of t.he rod contained between x and x+fl.x as illustrated in Fig. 1.2.1. If the thermal energy density is constant. throughoutthe volume, then the total energy in the slice is the product of the thermal energydensity and the volume. In general, the energy density is not constant. However, if

Q(x. t) = heat energy per unit volume generated per unit time,

4 Chapter 1. Heat Equation 1.2. Conduction of Heat in One-Dimension 5

perhaps due to chemical reactions or electrical heating. Q(x, t) is approximatelyconstant in space for a thin slice, and thus the total thermal energy generated perunit time in the thin slice is approximately Q(x, t)A Llx.

Conservation of heat energy (thin slice). The rate of change ofheat energy is due to thermal energy flmving across the boundaries and internalsources:

o x=a x= b L

Figure 1.2.2: Heat energy flowing intoand out of a finite segment of a rod.

1b

(De D1> )a Dt + Dx - Q dx = o.

This integral must be zero for arbitmry a and b; thE' area under the curve must bezero for arbitrary limits. This is possible only if t.he integrand itself is identicallyzero2 Thus, we rederive (1.2.3) as

1b D¢

1>(a, t) -1>(b, t) = - -dx,a AX

(t.his1 being valid if ¢ is cont.innously differentiable). Consequently,

if a and b are constants (and if e is continuous), This holds since inside the int.egralthe ordinary derivative now is taken keeping x fixed, and hence it must be replacedby a partial derivative. Every term in (1.2.4) is now an ordinary integral if we noticethat

(1.2.1)

(1.2.2)

:t [e(x, t)A Llxl "" 1>(x, t)A - 1'(". + Llx, t)A + Q(x, t)A Llx.

De _ I. ¢(x, t) - ¢(x + Llx, t) Q( )-D - 1m A + x, t ,

t t>.x--+O uxwhere the constant cross-sectional area has been cancelt:-d. Vie claim that thisresult is exact (with no small errorsL and hence we replacE' the ~ in (1.2.1) by = in(1.2.2). In this limiting process, Llx -4 0, t is being held fixed. Consequently, fromthe definition of a partial derivative,

Equation (1.2.1) is not precise because various quantities were assumed approxi­mately constant for the small cross-sectional slice. \Ve clailn that (1.2.1) becomesincreasingly accurate as 6x ----t O. Before giving a careful (and mathematically rigor­ous) derivation, we will just attempt to explain the basic ideas of the limit process,6x ---1- O. In the limit a') .6.x -----;. 0, (1.2.1) gives no interesting information, namely,o = O. However, if we first divide by Llx and then take the limit as Llx ~ 0, weobtain

De D¢-=--+QDt Dx .

(1.2.3) (1.2.5)

d 1b 1b

Dedt a edx= a otdx ,

Technically, an ordinary derivative d/dt appears in (1.2,4) since J: e dx dependsonly on t, not also on x. However,

Conservation of heat energy (exact). An alternative derivation ofconservation of heat energy has the advantage of our not being restricted to smallslices. The resulting approximate calculation of the limiting process (Llx ~ 0) isavoided. We consider any finite segment (from x = a to x = b) of the original one­dimensional rod (see Fig. 1.2.2). We will investigate the conservation of heat energy

in this region. The total heat energy is I: e(x, t)d:r, the sum of the contributions ofthe infinitesimal slices. Again it changes only due to heat energy flowing throughthe side edges (x = a and x = b) and heat energy generated inside the region, andthus (after canceling the constant A)

Equation (1.2.4), the integral conservation law, is more fundamental than thedifferential form (1.2.5). Equation (1.2.5) is valid in the usual case in which thephysical variables are continuous.

A further explanat.ion of the minus sign preceding D¢/Dx is in order. For exam­ple, if D¢/Dx > 0 for a <: x <: b, then the heat flux ¢ is an increasing function ofx. The heat is flowing greater to the right at x = b than at x = a (assuming thatb> a), Thus (neglecting any effects of sources Q), the heat energy must decreasebetween x = a and x = b, resulting in the minus sign in (1.2.5),

Temperature and specific heat. We usually describe mat.erials bytheir temperature,

IThis is one of the fundamental theorems of calculus.

2M~st proofs of this result are inelegant. Suppose that f(x) is continuous and J: f(x)dx = 0

for arbitrary a and b. We wish to prove f(x) = 0 for all x. We can prove this by assuming thatthere exists a point Xo such that f(xo) '# 0 and demonstrating a contradiction. If f(xo) '# 0 andf(x) is continuous, then there exists some region near XQ in which f(x) is of one sign. Pick a

and b to be in this region, and hence J: j(x)dx '# 0 since j(x) is of one sign throughout. This

contradicts the statement that J: j(x)dx = 0, and hence it is impossible for f(xo} '# O. Equation(1.2.5) follows.

(1.2,4)d lb 1b

- e dx = ¢(a, t) - ¢(b, t) + Q dx.&.a a

e(x, t)A t.x = c(x)u(x, t)pA t.x.

7

(1.2.7)

(12.6)

(1.2.8)

e(.r, t) = c(x)p(x)n(x, t).

au aq,c(x)p(.r)- = --I) + Q.at x

1.2. Conduction of Heat in One-Dimensiun

2. If thE'fe are temperature differences: the heat energy flows from the hotterregion to the colder region.

1. If the temperature is constant in a region, no heat energy flows.

This states that the thermal energy per unit volume equals the thermal energyper unit lIla,;;.., per unit degree tiInes the temperature times the matls density (massper unit volume). When the thermal energy density is elirninated using (1.:!.Ei).conservation of thermal energy, (1.2.3) or (1.2.5), becomes

4. The flow of heat, energy will vary for diffE'fent materials, even with the sametempentture differences.

In t.his way we have E'xplained the basic relationship bet\vepn thermal energy andtemperature:

3. The greater the ternpf'ratllrc differences (for the same material), the greateris the flow of heat energy.

Fourier's law. (Jsually, (1.2.7) is regarded as one equation in two unknowns:the temperature u(x, t) and the heat flux (flow per unit surface area per unit time)¢(x, t). How and \vhy does heat. energy flow? In other words, we need an expressionfor the dependence of the flow of heat energy on t.he temperature field. First wesummarize certain qualitative properties of heat flow with which we are all familiar:

Fourier (1768 - 1830) recognized properties 1 through 4 and summarized them (aswell as numerous experiments) by the formula

known as Fourier's law of heat conduction. Here 8u/Dx is the derivative ofthe temperature; it is the slope of the temperature (as a function of x for fixed t);it represents temperature differences (per unit length). Equation (1.2.8) stat.es thatthe heat flux is proportional to the temperature difference (per unit length). If thetemperature u increases as x increases (i.e., the temperature is hotter to the right),au/ax> 0, then we know (property 2) that heat energy flows to the left. Thisexplains the minus sign in (1.2.8).

Chapter 1. Heat Equation

u(x, t) = temperature,

specific heat (the heat energy that must be supplied to a unitmass of a substance to raise its temperature one unit).

c

In general, from experiments (and our definition) the specific heat c of a materialdepends on the temperature n. For example, the thermal energy necessary to raisea unit mass from DOC to 1°C could be different from that needed to raise the massfrom 85°C to 86°C for the same substance. Heat flow problems with the specificheat depending on the temperature are mathematically quite complicated. (Exer­cise 1.2.1 briefly discusses this situation.) Often for restricted temperature intervals,the specific heat is approximately independent of the temperature. Ho\veverJ exper­iments suggest that different materials require different amounts of thermal energyto heat up. Since WE' would like to formulate the correct equation in situationsin which the composition of our one- dimensional rod might vary from position toposition, the specific heat will depend on x, c = c(x). In many problems the rod ismade of one material (a uniform rod), in which CaEe we will let the specific heat cbe a constant. In fact, most of the solved problems in this te,,:t (as well as otherbooks) correspond to this approximation, c constant.

allowing it to vary with x, possibly due to the rod being composed of nonuniformmaterial. The total mass of the thin slice is pA t.x. The total thermal energy inany thin slice is thus c(x)u(x, t) . pA t.x, so that

Thermal energy. The thermal energy in a thin slice is e(x, t)A t.x. How­ever, it is also defined as the energy it takes to raise the temperature from a referencetemperature 0° to its actual temperature u(x, t). Since the specific heat is inde­pendent of temperature, the heat energy per unit mass is just c(x )u(x, t). We thusneed to introduce the mass density p(x):

p(x) = mass density (mass per unit volume),

not their thermal energy density. Distinguishing between the concepts of tempera­ture and thermal energy is not necessarily a trivial task. Only in the mid-170Gs didthe existence of accurate experimental apparatus enable physicists to recognize thatit may take different amounts of thermal energy to raise two different materials fromone temperature to another larger temperature. This necessitates the introductionof the specific heat (or heat capacity):

6

8 C;}mptcf 1. Heat Equation 1.2. Conduction of Heat in One-Dimension 9

Heat equation. If Fourier's law, (1.2.8), is substituted into the conservationuf heat cueq..,ry- equation, (1.2.7), a partial differential equation results:

We designate the coefficient of proportiona.Iity !f(j. It measures the ability of thematerial to conduct heat and is called the therIllal conductivity. Experimentsindicate that. different materials conduct heat Jifferf'nt.ly; [((I dCpCIlCh; on the partic­ular material The larger K o is~ the greater thf~ flow of heat energy with the sametemperature differences. A material with a low value of K o ,",,"ouJd be a poor con­ductor of heat energy (and ideally snited for hOII1C insulation). F(l[ a rod composedof different materials, K o will be a. function of :1'. Furthermore, experiments showthat the ability to conduct heat for most. materials is different at different temper­atures, Kn(x, 11,) . However- jll.st a.... \\'ith the specific heat (', t.he dependence on thetemperature itl often not important in part.icular problclIltl. Thutl, throughout thistext we will assume that t.he thermal conductivit.y' [{(1 only depends on x, Ko(x).Usually, in fHd. we "vill clitlcUSS uniform rods in \vhich [(0 is a constant.

iJ11. iJ ( iJU)ep-,- = -. K o- +q.iJ t iJx iJx

(1.2.9)

(such a..., perfumeH and pullutants) satisfi{~s the diffusion equation (1.2.8) in certainone - dimensional situatioIls.

Initial conditions. The partial differential equations describing the flow ofheat energy, (1.2.9) or (1.2.10), have one time derivative. When an ordinary differ­ential equation has one derivative l the initial vahle problem consists of solving thedifferential equation with one initial condit.ion. Newt.on's la\\' of motion for the posi­tion x of a particle yields a second-order orrliwu,\' difff'rential equation, md'2.r/dt2 =

forces. It involves second derivatives. The init.ial value problpIIl consist!; of solvingthe differential equation with two initial conditions, the initial position x and theinitial velocity d:r/dt. From these pieces of information (including the knowledge ofthe forces), by solving the differential pqnatiol1 with the initial conditions, we canpredict the future motion of {l particle in the x-diredion. We wish to do the sameprocess for our partial differential equation, that is. predict the future temperature.Since the heat equations have on(' time df'ri v<-1ti VE', we mUtlt. be given one initialcondition (Ie) (usually at t = 0), the initial temperature. It is possible that theinitial temperature is not constant, bnt depends on :1'. Thus, we must be given theinitial temperature distribution,

n(x, 0) = f(.I:).

If, in addition, there are no tlources, Q = 0, then after dividing by the constant cp,the partial diffprential equation becomes

iJn iJ'11.ep iJt = K oiJx' + q.

iJ 1"iJb " f(x)dx = f(b),

derive the heat equahon (1.2.9).

(a) Show that the heat energy per unit mass necessary to raise the temper­ature of a thin slke of thickness t.x from 0° to n(x, t) is not e(x)u(x, t),but instead Jo" c(x, u)du..

(b) R.ederive the heat equation in this case. Show that (1.2.3) remains un­changed.

EXERCISES 1.2

Is this enough information to predict the future temperature? \Ve know the initialtemperature diHtribution and that the temperature changes according to the partialdifferential equation (1.2.9) or (1.2.10). However, we need to know that happensat the two boundaries, J: = 0 and x = L. \\lithout knowing this informat.ion, wecannot predict the future. Two conditions are needed corresponding to the secondspatial derivatives present in (1.2.9) or (1.2.10), usually one condition at each end.We discuss these boundary conditions in the next section.

1.2.1. Suppose that the specific heat is a function of position and temperature,c(x, 11.).

1.2.2. Consider conservation of thermal energy (1.2.4) for an.y ,segment of a one­dimensional rod a. :::; x :s; b. By using the fundamental theorem of calculus

(1.2.10)(}U=k 02U.

iJ t iJ.r' .

where the constant k.

\Ve u:-:;ually think of the sources of heat energy Q as being given, and the onlyunknown being the temperature u(x, t). The thermal coefficients c, p, K n fill dependon the material and hence may be functions of x. In the special case of a uniform rod,in which c, p, K o are all constants, the partial differential equation (1.2.9) becomes

k = K o

ep

is called the thermal diffusivity, the thermal conductivity divided by the productof the specific heat and mass density. Equation (1.2.10) is often called the heatequation; it corresponds to no sources and constant thermal properties. If heatenergy is initially concentrated in one place, (1.2.10) will describe how the heatenergy spreads out, a physical process known a5 diffusion. Other physical quan­tities besides temperature smooth out in much the same manner, satisfying thesame partial differential equation (1.2.10). For this reason (1.2.10) is also knowna" the diffusion equation. For example, the concentration u(x, t) of chemicals

1.3 Boundary Conditions

1.2.4. Consider a thin one - dimensional rod without sources of thermal energ}',.....hose lateral surface area is not insulated.

Prescribed temperature. In certain situations, the temperature of theend of the rod, for example :r = l\ lllay be approximat.ed by a prescribed tenl­perature,

11

(1.3.3)

(1.3.4)

(1.3.5)AU

-Ka(L) ax (L, t) = H[u(L, I) -lIB(I)],

where ¢(t) is giV€Il. This is equivalent to giving one condition for the first derivative,Du/D.T, at .r = n. The slope is givc'll at :r = O. Equatioll (] .:3.2) cannot be integratedin :t becau~c the slope is knowlI onl.v at one value of ;1;. The simplest example of theprescribed heat flow boundary condition is when an end is perfectly insulated(somctiIIH-'-" we omit the "perfectly"). In thi:-> case there is no heat flow at theboundruy. If:1' = 0 is insulated. then

8'1L .-.(0,1) = a.ax

1.3. Dowldary (.'nndi tiOllS

3For another sit.uation in which (1.3.4) is valid, see Berg and McGregor [1966].

-](0(0) ~u (a, t) = -H[lI(O, t) -"n(t)l.u.T

where the proportionalit), constant H is called the heat transfer coefficient (orthe convection coefficient). This bOllndary condition:1 involves a linear comhinationof 11 and o11/8:r. vVe lllUHt bE' careful with the sign of proportionality. If the rodis hotter thml the bath [u(O. I) > UB(t)], then usually heat flows out of the rod atx = O. Thus, heat iH flowing to the left, and in this case the heat flow would benegative. That is why we introduced a minuo oign in (1.3.4) (wit.h H > 0). The sameconclusion would have been reached had we assumerl t.hat u(O, t) < 'UB(t). Anotherway to nnderst.and the signs in (1.3.4) is to again assnme that u(O, i) > UIJ(t). Thetemperature is hotter to the right at x = 0 and we should expect the temperatureto continue to increase to the right. Thus, au/ax should be positive at x = 0.Equation (1.3.4) is consistent with this argument. In Exercise 1.:).1 you are a.'3kedto derive, in the same manner, that the equation for Newton's law of cooling at aright end point :1' = L is

Newton's law of cooling. \Vhen a one - dimellsional rod is in contact atthe houndary ""ith a moving fluid (e.g.. air), then neither the prescribed temperaturenor t.he prc:->nibec! heat flow rnay be appropriate. For example) let ll~ imagine avery wann rod in contact with cooler moving air. Heat will leav{~ the rod) heat.ingup the air. Tht~ air will then carry the heat away. This process of heat transfer iscalled convection. However, the air will be hotter near the rod. Again, this is acomplicated problem: t.he air temperature will actually vary with distance from therod (ranging between t.he bath anc! rod temperatures). Experiments show that, as agood approximation, tIlE' l1l'at flow leaving the rod is proportional t.o the temperaturedifferellce between the bar and the pl'e:-:;cribcd external temperature. This boundaryconditioll is called Newton's law of cooling. If it is valid at ;r = 0, then

where UB(t) is the external tetuperature at x = L. We immediatel~' note thesignificant. sign difference between the left boundary (1.3.4) and the right boundary(1.3..5).

The coefficient H in Newton's law of cooling is experimentally determined. Itdepends on properties of the rod as well as fluid propert.ies (including the fluid

(1.3.2)

(1.2.11)au a ( . 011) Pcp- = - K o-.- - -[u(J', t) - 'Y(:IO, t)]h(x),at OJ: ax A

where h(x) is a positive x-dependent proportionality, P is the lateralperimeter, and A is the cross-sectional area.

Compare (1.2.11) to the equatioll for a one - diml~ll~ional rod whose lat­eral surfaces are insulated, hut ",'ithout heat soun·l;-'~.

Specialize (1.2.11) to a rod of circular cross section with COll~tant thermalproperties and 0° outside kmperature.

Consider the assumption~ in part (d). Suppose that the temperatllrein the rod is uniform [i.e., u(x. t) = u(i)]. Determine u(t) if initiallyufO) = 110'

As;.;urne that the heat energy flowing out of the lateral :-:lide:-:l per unit. ~ur­face area per unit time is w(x, t). Derive the partial differential equationfor the temperat.ure u(x, t).

Assume that w(x, t) is proportional to the klllperature difference be­tween the rod u(:r, t) and a known outside temperature "·y(x, t). Derivet.hat

(e)

(a)

(d)

(b)

*(e)

Insulated boundary. In other situations it is possible to preocribe t.heheat flow rather than the temperature,

au-Ko(O) ax (0, t) = <I>(t),

In solving the heat equation, either (1.2.9) or (1.'2.10), one boundary condition(Be) is needed at each cud of t.he rod. The appropriate condition depends on t.hephYHical mechanism in effect at each end. Oft.en the condition at the boundar}'depends on both the material inside and olltHide the rod. To avoid a l110re difficultmathematical problem, we ,,,,ill aHHUllW t.hat the outside Pllvironment is known, notoignificantly alt.ered by the rod.

u(O, t) = UB(t). (1.3.1)

where UB(t) io the temperatlll'e of a f1nid bat.h (or reservoir) with which t.he rod isin contact.

10 Chapter 1. Heat Equation

/1.2.3. If u(x, t) is known. give an expression for the total thermal pnergy containedin a rod (O<x<L).

12 C~hapter 1. Heat Equation 1.4. bqwl1brium 'l(lmperature lJistriiJlltioJl 13

velocity). If the coefficient is very small, then very little heat energy flows acrossthe boundary. In the limit a~ H ---+ 0, Ne\vton's law of cooling approaches theinsulated boundary condition. vVe can think of Newton\:; law of cooling for H # 0 at;representing an imperfectly insulated boundary. If Ii ---+ ::0, the boundary conditionapproaches the one for prescribed temperature, u(lL t) = un(t). This i:-:; most ea."ily'seen by dividing (1.3.4), for example, by H:

*1.3.3. Consider a bath containing a fluid of specific heat ('f and mass density Pfwhich surrounds t.l1P end x = L of a one - dimPll:-iional rod. Suppose that thebatb is rapidly' stirred in a manner such that. the bath tl?lI1peratllre i~ Clp­proximately uniform throughout, equaling the temperat.ure at .1' = L, u(L, i).Asmme that the bath it; thermally insulated except at its perfect thermalcontact with the rod, where the bath Ill<":\' be heated or cooled by the rod.Determine all equation for the tempen1Jure in the bath. (This will he ab"undary condition at the end:c = L,) (Hint: Sec Exercise L3,2,)

Thus, H ---+ 00 corresponds to no insulation at all, 1.4 Equilibrium Temperature Distribution

Summary. We have described three different kinds of boundary conditions.For example, at x = 0:

ufO, I)

-Ko(0)1)~(0,t)

- K (0)"" (0 I)u {'x '

us(t) prescribed temperature

¢(t) prescribed heat flux

-Hlu(O, I) - "B(t)] Newton's law of cooling

1.4.1 Prescribed Temperature

Let us now fOrIIHllate a simple, hut ty'pical, problem of heat fiow. If the ther­mal coefficients are constant and there are 1l<J sources of thprmal energy. then thet.E:'.rnperature u(x, t) in a one- dimensional rod 0 S; :r ~ L satisfies

(lA1I

The solutiou of this partial differential eqwltiol1 must satisf}T the initi"l condition

and oue boundary condition at each end. For example, each end might be in conta.ctwith different large baths, such that the temperature at each entl. is prescribed

(U4)

(14,;))

(U2)

and

~J~

,,(x, 0) = f(:r)

u(O, I)U(L, I)

ufO, t) = T,

Equilibrium temperature distribution. Before we begin to attacksueh an initial and boundary value problem for partial differential equations, W(~

discuss a physically related question for ordinary differential equations. Suppm;pthat the boundary conditions at x = 0 and .r = L were steady (i.e .. independentof time),

One aim of this text is to enable the reader to solve the problem sp<-'Cified by (1.4.1- 1,4:3).

where T] and T2 are given constants. vVe define an equilibriunl or steady-statesolution to be a temperature dist,ribution that <-loeH not depend on time, that is,u(x, I) = u(x), Since a/al u(x) = 0, the partial differential equation becomesk(8 2 u./8x2

) = 0, but partiRI derivatives arC' Hot necessa.ry, and thus

u(O, t) = 100 - 25 cos I,

and for the right end, x = L, to be insulated,

and no heat energy is lost at x = Xo (i.e., the heat energy flowing out ofone flows into the other). What mat.hematical equation represents the lattercondition at x = xo? Under what special condition is au/ax continuous atx = xo?

u(xo-, I) = u(.[o+, I)

These same conditions could hold at x = L, noting that the change of sign (-Hbecoming Ii) is necessary for Newton's law of cooling. One boundary conditionoccurs at each boundar)'. It is not necessary that both boundaTip;.; ;.;atisfy the samekind of boundary condition. For example, it is possible for x = {) to have a prescribedoscillating temperature

EXERCISES 1.3

1.3.1. Consider a one-dimensional rod, 0::; x ::; L. Assump that the heat energyV flowing out of the rod at x = L is proportional to the temperature difference

between the end tempE'rature of the bar and the knmvn external t.emperature.Derive (1.3,5) (briefly, physically explain why H > 0),

$' *1.3.2. Two one - dimensional rods of different materials joined at 2' = Xu are saidto be in perf€ct thermal contact if the temperature is continuous at x = xo:

14 Chapter 1. Heat b'quatlOIJ 1.4. l!..'quilibrium 'lemperature Distribution 15

The boundary conditions are distribution, since the boundary conditions are independent of time:

In doing steady-st.ate calculations, the initial conditions are usually ignored. Eqlla­tion (1.4.4) is a. rather trivial second-order ordinary differential equation. Its gen­eral solution may be obtained by integrating twice. Integrating (1.4.4) yieldsdu/dx = C 1 , and int.egrating a second time shows that 1.4.2 Insulated Boundaries

In Sec. 7.2 we will ,olve the time-dependent problem and show that (1.4.9) issatisfied. However, if a .steady state is approached, it is morp pasil)' obtained bydirectly solving the equilibrium problem.

u(O) = T 1

u(L) = T 2 .(1.4.5)

. ( ( T2 - T jhmux,I)=ux)=T, + x.

t--+CXi L ( lA.9)

(1.4.14)

(1.4.15)

(1.4.16)

~u(O) = 0(X

du .: -(L) = 0

dx 'BC2

BCl

082uODE: -. =0

dxz

PDE:au alItf) I = k o.r' (1.4.10)

IC: U(.L 0) = f(x) (1.4.11)

BCl:f)u~(O, t) = [) (1.4.12)x .

BC2:auax (L,I) = O. (1.4.13)

As a second example of a steady-state calculation, we consider a one - dimensionalrod again with no sources and with constant thermal properties, but this time withinsulated boundaries at x = 0 and x = L. The formulation of the time - dependentproblem is

The equilibrium problem is derived by setting a/a I = O. The equilibrium tempera­ture distribution satisfies

u = C,x + C2 . (1.4.17)

The boundary conditions imply that the slope must be zero at both ends. Geo­metrically, any straight line that is fiat (zero slope) will satisfy (1.4.15, 1.4.16) as

where the initial condition is neglected (for the moment). The general solution ofd2u/ih;2 = 0 is again an arbitrary straight line,

(1.4.8)

(1.4.7)

(1.4.6)

Figure 1.4.1: EquilibriuIll tempera­ture distribution.

implies T1 = C2

implies Tz = C,L + C2 .

x=L

u(O) = T j

u(L) = Tz

x=O

It is easy to solve (1.4.7) for the constants C2 = T j and C, = (T2 - TJl/ L. Thus,the unique equilibrium solution for the steady-state heat equation with these fixedboundary conditions is

Approach to equilibrium. For the time- dependent problem, (1.4.1) and(1.4.2), with steady boundary conditions (1.4.5), we expect the temperature distri­bution u(x, t) to change in time; it will not remain equal to its initial distributionf(;-r). If ·we wait a very, very long time, we would imagine that the influence of thetwo ends should dominate. The initial conditions are usually forgotten. Eventually,the temperature is physically expected to approach the equilibrium temperature

u(x)

We recognize (1.4.6) as the general equation of a straight line. Thus, froIll theboundary conditions (1.4.5) the equilibrium temperature distribution is the straightline that equals T j at x = 0 and Tz at x = L. as sketched in Fig. 1.4.1. Geomet­rically there is a unique equilibrium solution for this problem. Algebraically, wecan determine the two arbitrary constant.s, C 1 ami Cz, hy appl)'ing the boundaryconditions, u(O) = T j and u(L) = T2 :

illustrated in Fig. 1.4.2. The solution is any constant temperature. Algebraically,from (1.4.17), dujdx = C j and both boundary conditions imply C j = o. Thus,

•

17

* (a) Q =0, 11(0) = 0, u(L) = T

(b) Q =0, u(O) = T, alL) = 0

(c)au

1I(L)=TQ =0, 9(0) = 0,ex

au* (d) Q =0, 11.(0) = T, -(L) =0'

ax

(e) ~=1 ,,(0) = T j , 1I(L)=T2K o '

* (I) ~ =x' u(O) = T, aU(L) = °K o ' ax

(g) Q =0, 11(0) = T, ~.~ (L) + u(L) = 0

* (h) Q =0, ~: (0) - [11(0) - TJ = 0, au (L) ="ox

1.4. Equilibrium Temperature Distribution

*(a) Determine the heat energy generated per unit time inside the entire rod.

(b) Determine the heat energy flowing out of the rod per unit time at x = 0and at x = L.

(c) What relationships should exist between tbe answers in parts (a) and(b)?

/1.4.3. Determine the equilibrium temperature distribution for a one-dimensionalrod composed of two different materials in perfect thermal contact at x = 1.For 0 < x < 1, there is one material (cp = 1, K o = 1) with a constant source(Q = 1), whereas for the other 1 < x < 2 there are no sources (Q = 0, cp =2, Ko = 2) (see Exercise 1.3.2) with 11(0) = 0 and u(2) = o.

J1.4.4. If both ends of a rod are insulated, derive from the partial differential equa­tion that the total thermal energy in the rod is constant.

the average of the initial temperature distribution. It is as though the initialcondition is not entirely forgotten. Later we will find a u(x, t) that satisfies (1.4.10- 1.4.13) and show that lilllt _ x , "(1·, t) is given by (1.4.21).

In these you may assume that ,,(.r, 0) = f(l·).

\/"1.4.2. Consider the equilibrium temperature distribution for a uniform one- dimensionalrod with sources Q/ K o = x of thermal energy, subject to the boundary con­ditions u(O) = 0 and u(L) = O.

EXERCISES 1.4

/1.4.1. Determine the equilibrium temperature distribution for a one-dimensionalrod \vith constant thermal properties with the follO\\'ing sources and boundaryconditions:

(1.4.19)

(14.21)

(1.4.20)

(1.4.18)

Chapter 1. Heat Equation

Figure 1.4.2: Various constant equi­librium temperature distributions(with insulated ends).

1I(X) =C,

x=L

1 rL

u(x) = C2 = L Jo f(x) dx,

x=o

d rL au a1ldt Jo cpu dx = -Koax (0, t) + K oax (L, t).

Since both ends are insulated,

1L

cpu dx = constant.

One implication of (1.4.20) is that the initial thermal energy must equal the fi­

nal (limt_=) thermal energy. The initial thermal energy is cp JoLf(x)dx since

u(x,O) = f(x), while the equilibrium thermal energy is cp JoL C2dx = cpC2 L sincethe equilibrium temperature distribution is a constant u(x ,t) = O2 . The constantO2 is determined by equating these two expressions for the constant total ther­mal energy, cp j~L f(x)dl: = cpC2L. Solving for C2 shows that the desired uniquesteady-state solution should be

if we wait long enough a rod with insulated ends should approach a constant tem­perature. This seems physically quite reasonable. However, it does not make sensethat the solution should approach an arbitrary constant; we ought to know whatconstant it approaches. In this case. the lack of uniqueness was caused by the com­plete neglect of the initial condition. In general , the equilibrium solution will notsatisfy the initial condition. However, the particular constant equilibrium solutionis determined by considering the initial condition for the time- dependent prob­lem (1.4.11). Since both ends are insulated, the total thermal energy is constant.This follows from the integral conservation of thermal energy of the entire rod [see(1.24)1:

for any constant C2 . Unlike the first example (with fixed temperatures at bothends), here there is not a unique equilibrium temperature. Any constant tempera­ture is an equilibrium temperature distribution for insulated boundary conditions.Thus, for the time - dependent initial value problem, we expect

lim u(x, t) = C2 ;t~=

16

jwhere the heat energy within an aTbitrary subregion R is

simpler ones) so that when we do di~("uss techniques for the .:;olutions of PDEs, wewill have Illore than one example to \\tork with.

heat energy = ffI cpu dV,

R

1!l

heat energy generatedinside per unit time,_ -...

heat energy flowingacross the boundaries +per unit time

rate of changeof heat energy

l.b. Heat .t!.·quatlOIl 111 Two OT '1 'ilree 1JlllWI1SiOllS

Heat energy. We begin our derivation by considering any arbitrary subre.glOuR a.s illustrated in Fig. 1.5.1. As in the one- dimensional ease. conservation of lWiitenergy is summarized by the following word equation:

18 Chapter 1. Heat t;quatioll

/1.4.5. Considel a oue-dimensional rod 0 < x < L of knmvn length and knowncon.:;tant thermal properties without sources. Suppu::;e that the temperatureis an unknown constant T at x = L. Determine T if v·/e know (in the steadystate) both the temperature and the heat Row at x = O.

1.4.6. The two ends of a uniform rod of length L are insulated. There i::; a constantsource of thermal energy Qu i- 0 and the temperature b initiall.y u(.r. 0) =

f(x).

(a) Show mathematically that there does not exbt any equilibrium temper­ature distribution. Briefl)' explain physically.

(b) Calculate the total thermal energy in the entire rod.

1.4.7. For the following problems, determine an equilibrium tempf'rature distri­bution (if one exists). For what values of ,13 aI'(=' t.lwre solutions? Explainphy,ically.

Introduction. In Sec. 1.2 we showed that for the conduction of heat in aone-dimensional rod the temperature u(x, t) satisfies

...

Figure 1.5.1: three- diInensiollal suhregion R.

Heat flux vector and nonnal vectors. We need an expression forthe flow of heat energy. In a one - dimensional problem the heat Rux ¢ is definedto the right (q) < 0 means flowing to the left). In a three-dimel"ional problemthe heat flows in some direction, and hence the heat flux is a vector 4>. Themagnitude of 4> is the amount of heat energy Rowing per unit time per unit snrfacearea. However) in considering conservation of heat energy it is oIlI~v the heat Rowingacross the. bO'llndaries per unit time that is important. If, a." at point A in Fig. 1.5.2.the heat flow is parallel to the boundary, then there is no heat ellergy cTO.'Jsing th~boundary at that point. In fact, it is only the normal component of the heat flowthat contributes (as illustrated by point B in Fig. 1.5.2). At any point there are twonormal vectors, an inward and an outward normal n. \Ve will use the cOIlventionof only utilizing the unit outward normal vector n (where the' stands for aunit vector).

Conservation of heat energy. At each point the amount of heat energyfiowmg out of the region R per unit time per unit surface area is the outward normalcomponent of the heat flux vector. From Fig. 1.5.2 at point B, the outward normalcomponent of the heat flux vector is 14>1 cos e= 4> •n/lnl = 4> . n . If the heat flux~ector 4> is directed inward, then 4> • n< 0 and the outward flow of heat energyIS nega.tive. To calculate the total heat energy flowing out of R per unit timf'. we

instead of the one dilnensional integral used in Sec. 1.2.iJu iJ'u iJu an (

*(a) ---+1 n(x,O) = fIx), a(o. I) = 1. ax L,I)=(3iJl - ax' ' l' .

au a2uu(:!',O) = f(x),

an iJn(b)

iJt=Ux" ax (0, t) = 1, -(L, t) = (3a:r

an iJ'n a·u Uu(e) 3t = iJx' +:C - 13. u(x,O) = f(ll ax (0, t) = 0, a(L.I) = 0

~r

where k = K o/ cpo Before we solve problems involving these partial differentialequations. we will formulate partial differential equations corresponding to heat flowproblems in two or three spatial dimensions. We will find the derivation to be similarto the one used for one - dimensional problems, although important differences willemerge. We propose to derive new and more complex equations (before solving the

In case~ in which there are no sources (Q = 0) and the thermal properties areconstanL the part.ial differential equation becomes

1.5 Derivation of the Heat Equationin Two or Three Dimensions

au a ( . au)cp-u. = -,- 1\0-.- + Q.t Ux iJx

..

20 Chapter 1. Heat. nquatlOlJ 1.0. Heat .t,:quatiun in Two or T1Jn-'t' Dillwllsions 21

BA

JI/\7.A dV ~.1fi A· n d8.R

( 1.5.3)

Figure 1.5.2: Outward normal component of heat flux vector.

(1.5.4)

(1.5..5)

(1.5.6)au

CPat = -\7.</>+ Q.

:t1 cpu dV = -I \7.</> dV +1 Q dVR R R

1JJ[cP~~+\7'</>-Q] dV=O.R

Since this integral is zero for all regions R, it follows (as it did for one-dimensionalintegrals) that the integrand itself must be zero:

We note that the time derivative in (1.5.4) can be put inside the integral (since Ris fixed in space) if the time derivative iH changed to a partial derivative. Thus, allthe expressions in (1.5.4) are volume integrals over the Harne volume, and they canbe combined into one integral:

This is also known as Gauss's theorem. It can be used to relate certain surfaceintegrals to volume integrals, and vice versa. It is very important and very useful(both innnediately and later in this text). We omit a derivation, which may be basedon repeating the one - dimensional fundamental theorem in all three dimensions.

Application of the divergence theorem to heat flow. In partic­ular, the closed surface integral that arises in the conservation of heat energy (1.5.1),corresponding to the heat energy flowing across the boundary per unit time, canbe written as a volume integral according to the divergence theorem, (1.5.3). Thus,(1.5.1) becomes

aucp at + \7.</>- Q = 0

Of, equivalentI;y,

Equation (1.5.6) reduces to (1.2.3) in the one- dimensional case.

Fourier '8 law of heat conduction. In one - dimensional problems,from experiments according to Fourier's law, the heat flux ¢ is proportional to thederivative of the temperature, ¢ = -Koau/ax. The minus sign is related to thefact that thermal energy flows from hot to cold. au/ax is the change in temperatureper unit length. These same ideas are valid in three dimensions. In an appendix,we derive that the heat flux vector </J is proportional to the temperature gradient

( \7u = 8ui + i1..!!.1~ + aUk).- ax By 8% •

(1..5.2)

(1.5.1)

a a a\7·A"" -Ax + -aAy + -aA,.ax y z

Note that the divergence of a vector is a scalar. The divergence,theorem s~ates

that the volume integral of the divergence of any contmuously dIffer­entiable vector A is the closed surface integral of the outward normalcomponent of A :

that is the flow through the boundaries can be expressed as an integral over theentire ;egion for one - dimensional problems. We claim that the d~vergence theoremis an analogous procedure for functions of three V'<1riable~. The dlver~ence ~heore!U

deals with a vector A (with components Ax> Ay and A,; I.e., A = Axt+Ay]+A,k)and its divergence defined as follows:

4Sometimes the notation ¢n is used instead of 4J • fl., meaning the outward normal componentof ¢ .

Divergence theorem. In one dimension, a way in which we derived apartial differential relationship from the integral conservation law was to notice(via the fundamental theorem of calculus) that

l "a¢¢(a) - ¢(b) = - a ax dx;

must multiply </J ' n by the differential surface area dB and "surn" over th: entiresurface that encloses the region R. This4 is indicated by the closed surface mtegTal~ ¢ . n dS. This is the amount of hea:t energy (per unit time) leavi~lg .the regionR and (if positive) results in a decreasmg of the total heat euergy wlthm R. If Qis the rate of heat energy generated per unit volume, then the total heat energygenerated per unit time is JfJR Q dV. Consequently, com;ervation of heat energyfor an arbitrary three - dimensional region R becomes

•

I'1

Uhapter 1. Heat Bquation 1.5. Heat Eqlla{.ion in Two or Three Dimensions 23

(1.5.7)!1=-KOVU,\

known as Fourier's law of heat conduction, where again K o is called the thermalconductivity. Thus, in three dimensions the gradient Vu replaces au/ax.

Heat equation. When the heat flm, vector, (1.5.7), is substituted into theconservation of heat energy equation, (1.5.6), a partial differential equation for thetemperature results:

Initial boundary value problem. In addition to (1.5.8) or (1.5.11),the temperature sH.tisfies a given initial di.stributioll.

u(:r, y, z, 0) = i(":. y. 0).

The temIH:'rature also Hatisfies a boundary condition at every point on the surfacethat encloses tIK' region of interest. The boundary condition call be of various types(as in the one-dimensional problem). TIlE' temperature could be prescribed,

u(x, y, z, t) = T( x. y, Z, t),

(1.5.12)-J(oVu·il = H(u -11,,).

ever,vwhere on the bonndary w}wrE' T is a known functiun of t at each point ofthe boundary. It. is also possible that the flow across the boundary is prescribed.Frequently, we might have the boundary (or part of the boundary) insulated. Thismeans that there is no heat flow aeros.s that portion of the boundary. Since theheat flux vector is - J("o \In, the heat flowing out wjJl be the unit outward normalcomponent of the heat flow vectoL - K o\lu·ii, where ii b (l unit outward normalto the boundary surface. TIlllS, at an insulated surface,

Note that usually the proportionality constant H > 0, since if u > HI)] then we ex­pect that heat energy will flow out and - J(o '\7u·ii will be greater thall zero. Equa­tion (1.5.12) verifiE's the two forms of Newton's law of cooling for one-dimensionalproblems. In particular, at x = 0, il = -i and the left-hand ,ide (1.h.8.) of (1.0.12)becomes KOal1/aX, while at :e = L, il = i and the 1.h.s. of (1.5.12) hecomes-Koa,,/ax [Dee (1.3.4) and (1.3 ..5)].

Recall that \lu·ii is the directional d(~rivative of 'U in the ontward normal direction:it is also called the normal deriv'dtive.;3

Often Newton '8 law of cooling lH it more realistic condition at the boundar)'.It states that thE' heat energy flowing out per unit time per unit Hnrface area isproportional to the difference between the t,cll1pcrature at thl' surface ~{ and thetemperature outside the surface U/;o Thus, if Newton's Imv of cooling b valid, thenat the boundary

Vu·il = o.

0= V·(KoV,,) +Q.

Steady state. If the boundary conditiolls and any sources of thermal energyare independent of time, it is possible that there exist steadY-Htate solutions to theheat equation satis(ying t.he given steady boundary condition:

5Sometimes (in other books and references) the notation a ulan is used. However, to calculateaulan one usually calculates the dot product of the two vectors, Vu and ii. Vu,n, so we will notuse the notation aulan in this text.

Note that an equilibrium temperature distribution u(x, y, z) satisfies a partial differ­ential equation when more than one spatial dimension is involved. In the case with

(1.5.9)

(1.5.8)

(1.5.11)a" _ k,",2- v u.at

au- = kV·(Vu),at

epau = V.(KoVu) + Q.at

a, a, a.v =0 ax' + ayJ + az k.

Note that V" is V operating on ", while V·A is the vector dot product of del withA. Furthermore, V 2 is the dot product of the del operator with itself or

Equation (1.5.11) is often known as the heat or diffusion equation in three spatialdimensions. The notation \72u is often used to emphasize the role of the del operator

V:

a (au) a (au) a (au)V·(Vu) = a,. a.r + ay ay + az az =

operating on u, hence the notation del squared, \72.

(1.5.10)This expression \72 u is defined to be the Laplacian of u, Thus, in this case

where k = Ko/ep is again called the thermal diffusivity. From their definitions, wecalculate the divergence of the gradient of u:

In the cases in which there are no SOllices of heat energy (Q = 0) and the thermalcoefficients are constant, (1.5.8) becomes

24 Chapter 1. Heat b'quation 1.5. Heat Equation in TWfJ or Three Dimensions 25

constant thermal properties, the equilibrium temperature distribution will satisfy Polar and cylindrical coordinates. The Laplacian.

(1.5.13)(1.5.17)

Sometimes (1.5.16) is called Green's theorem, but we prefer to refer to it as tbe two­dimensional divergence theorem. In this way only one equation need be familiar tothe reader, narnely (1.5.15); the conversion to two - dimensional form involves onlychanging the number of integral signs.

(1.5.18)

(1.5.19)

(1..5.20)

rcosersine

z z,y

V2u = ~~ (r au) + a2urar ar az2.

V2u = ~~ (r aU )rar 8,.

the Laplacian can be shown to equal t,he following formula:

There may be no Heed to memorize this formula, as it can often be looked up in areference book. As an aid in minimizing errors, it should he noted that every termin the Laplacian has the dimension of u divided by two spatial dimensions Uust asin Cartesian coordinates, (1.5.17)]. Since e is measured in radians, which have nodimensions, this remark aids in remembering to divide 8 2ulae2 by r2 . In polarcoordinates (by which we mean a two - dimensional coordinate system with.:: fixed,usually z = 0), the Laplacian is the same as (1.5.19) with iJ'u/az2= 0 since thereis no dependence on z. Equation (1.5.19) can be derived (see the exercises) usingthe chain rule for partial derivatives, applicable for changes of variables.

In some physical situations it is known that the temperature does not dependon the polar angle e; it is said to be circularly or axially symmetric. In thatcase

is important for the heat. equation (1.5.11) and its steady-state vcrsion (1.5.14), aswell as for other significant problem:,:, in science and engineering. Equation (1.5.17)written as above in Cart.esian coordinates is most useful when the geometrical regionunder investigation is a rectangle or a rectangular box. Other coordinate systemsare frequently llsefuL In practical applications, one may need the formula thatexpresses the Laplacian in the appropriate coordinate system. In circular cylindricalcoordinates, with r the radial distance from the z-axis and () the angle

Spherical coordinates. Geophysical problems as well as electrical prob­lems with spherical conductors are best solved using spherical coordinates (p, e, ¢).The radial distance is p, the angle from the pole (z-axis) is ¢, and the cylindrical(or azimut.hal) angle is e. Note that if p is constant and the angle ¢ is a constant a

(1.5.16)

(1.5.15)

(1.5.14)

If V·A dS = fA' it dT.R

Jf/ V·A dV = iJA. it dS.

R

is valid in two dimensions, taking the form

2 a2u a2uV u = ax2 + a y2 = 0,

since 8 2u18::;2 = O. two- dimensional results can be derived directly (without takinga limit of three- dimensional problems), by using fundamental principles in twodimensions. We will not repeat the derivation. HO\,.lever, we can easily outlinE' theresults. Every time a volume integra,} (jJfn ... dV) appears, it must be replaced by asurface integral over the entire two-dimensional plane region (JR'" dB). Silnilarly,the boundary contribution for three - dimensional problems. \vhich is the closedsurface integral 11> ... dS, must be replaced by the closed line integral f ... dT, anintegration over the boundary of the two - dimensional plane snrface. These resultsare not difficult to derive since the divergence theorPIIl in three dimensions,

Two - dinlensional problems. All the previous remarks about thrce­dimensional problems are valid if the geometry is such that the temperature onlydepends on x, y and t. For example, Laplace's equation in two dimensions, x andy, corresponding to equilibrium heat flow with no sources (and COIlst.ant thermal

properties) is

the Laplacian of the temperature distribution is .,;ero. Equation (1.5.14) is knuwnas Laplace's equation. It is also known as the potential equation, since thegravitational and electrostatic pot.entials satisfy (1.5.14) if there are no sources. v'lewill solve a number of problems involving Laplace's equation in later sections.

IV2u = 0; I

known as Poisson's equation.If, in addition, there are no sources (Q = 0), then

circle is generated with radius psin¢ (as shown in Fig. 1.5.3) so t.hat. Consider the polar cuordinates

I

I I

II

26 Chapter 1. Heat Equation 1.5. Hellt Equation in Two or Three DirnensiOllH

vG.3.27

The angle from the pole ranges from () to rr (v,..hilc the usual c.vlinclrical angle rangesfrom 0 to 211"). It can be shown that the Laplacian satisfies

1.5.1. Let c(x, y, z, t) df'note the concentration of a pollutant (the amollnt per unitvolume).

and{lr = sin B. Be = case2t/ By r'

:D = rcos()

y = r..,in().

Since 1'2 = x 2 + y2. show that (:,. r = co.., fJ., ' <.:riJe _ -~ille

D.I: - l'

(a)

*(a) if the outer radil.L5 is at temperature Tz and the inner at T1

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(b) if the outer radius is insulated and the inner radius is at temperature T1 .

,I' 'I7u·fi dB = a

(it) Show that the total heat energy is 211" J: cpur dr.

(h) Show that the flow of heat energy per unit time out of the annulus atr = b is -211"bKooujo,. Ir~b. A similar resnlt holds at T = a.

(c) Use parts (it) and (b) to derive the circularly symmetric heat equationwithout sources:

(b) Show that l' = cos oi + sin 0] and 0 = - sin oi + cos 0].(c) Using the chain rule, show that. '1'11 = %"<'i" + ~~~9.

(d) If A = A,.f+ A80, show that 'I7·A = ~ ;" (rA") + ~ ,;~ (AoJ. since 81' j 88 =

oand iJO j 80 = -1' follows from part (b).

( ) Show that '12 ,u = lJL (r~) + I i)'11/.e 1'81' <.1r ~~

0.5.4. Using Exercise 1.5.3(a) rind the chain rule for partial derivatives. derive thespecial case of Exercise 1.5.3(c) if u(I') OIlI}·.

11.5.5. AssUl~le th,at the temperature is drcnlarly s}·mnH:'tric. 11 = u(1', t), whereV - 1'2 = x 2 + y2. \Ve will derive the heat eqnation for this problem. Consider

allY C'ircular annulus a. :S r ::S b.

for any closed surfare. (Hint: Use the divergence theorem.) Give a ph)'sical/ interpretation of this re~;nlt (in the context of heat flow).

\/"1.5.9. Determine the equilibrium tE'lllperature distribution inside a circular annulus(r1 S r S r2)'

8u = ~.!!- (r 8U)iJt ,. ar Or .

lvI'.'dify Exercise 1.5.5 if the thermal properties depend on r.

1.5.7. Derive the heat equation in two dimensions by using Green's theoreHll

(1.5.1G),the two - dimensional form of the divergence theorem.

',0'5.8. If Laplace's equation is satisfied in three dimensions, show that

(1.5.22)

(1.5.21)p sin dJ cos ()psin ¢ sin e(J cos 1)·

yz

x

What is an expressiull for the tuta.I amount of pollutant in the region R?

Suppose that the flow J of the pollutant is proportional to the gradientof the concentratioll. (Is this rea.50nable'?) Express conservation of thepollutant.

(c) Derive the partial differential equation governing the diffusion of thepollutant.

(a)

(b)

:~ ppsin¢ _

--iJ

x=L

Figure 1.5.3: Spherical coordinates.

~"magnified

? 1 0 ( ?ou) 1 8 ( . .ou) 1 iJ'lI'I7-u = 2""'-) Vc;- + ,. . c;" sm<p-;:;-:- + 2 . '~. Dn"p (p vp P sm<p V<y v'!J p 8m <y v

*1.5.2. For conduction of thermal energy, the heat flux vector is <p = - KoVu.If in addition the molecules Hlove at an average velocity V a process calledconvection, then briefly explain why </J = -Ko'l7u + cpuV. Derive the cor­responding equation for heat flow) including both conduction and convectionof thermal energy (assuming constant thermal properties with no sources).

EXERCISES 1.5

28 Chapter 1. Heat Equation 1.5. Refit Equation in Two or Three Dimensions 29

(1.5.26)

(1.5.24)

av.- au- au­\1tl =' -i +-j + -k,

ax ay az

1 [ a a () 'J\1.p= I h.h -a (h"h",p")+-a. (h"h",p")+,.,---(h"h,.p,,.)tu l' "I/} 11 V u til

An expression for the divergence is more difficult to derive, and we ·will jw;tstate that if a vpd,or p is expressed in terms of this new coordinate s:rstemp = Pueu + !Jl,e l, + Pwew, then the divergence satisfies

6Examples of nonisotropic materials are certain crystaJ and grainy woods.

Appendix to 1.5: Review of Gradient and aDerivation of Fourier's Law of Heat Conduction

. t:.u au au aubm ~ = ai-a +"2- +"3- = a·\7u,

.6..,~---+O uS X By 8z

where it has been convenient to define the following vector:

au au aut:.u = u(x + t:.x, t) - u(x, t) '" -at:.x + "t:.y + -t:.z.x uy a2

Experimentally, for isotropic6 materials (i.e., without preferential directions) heatflows from hot to cold in the direction in which temperature differencesare greatest. The heat flow is proportional (with proportionality C'"on~tant K o, thethermal conductivity) to the rate of change of temperature in t.his direction.

The change in the temperature ~u is

In the direction & = 01 ~ + 0'2} +D:3k, ~x = .6..80, where ~s is the distance between:z: and x + Llx. Thus, the ratE' of change of the temperature in the direction 0: isthe directional derivative:

called the gradient of the temperature. From the property of dot products, if ais the angle between & and '\lu, then the directional derivative is I''\lul cos () sillce

1.5.19. Using (1.5.25), derive the Laplacian for cylindrical coordinates.

1.5.20. Using (1.5.25), derive the Laplacian for spherical coordinates.

1.5.18. Using (1.5.23) and, (1.5.24) derive the Laplacian in an orthogonal clII"Vi­linear coordinate system:

(1.5.23)

a.<r<bau = ~.!!..- (r au)at rar aT

au auv(r,O) = f(r), -(a,t) =(3, and -a(b,t) = 1.

ar r

Using physical reasoning, for what value(s) of (3 does an equilibrium temper­ature distribution exil->t?

subject to

1.5.15. Determine the scalt~ factors for cylindrical coordinates.

1.5.16. Determine the scale factors for spherical coordinates.

1.5.17. The gradient of a scalar can be expressed in terms of the new coordinatesystem \1g = aar/au + bar/a v + car/a w, where you will determine thescalars a, b, c. Using dg = '\l9 . dr, derive that the gradient in an orthogonalcurvilinear coordinate system is given by

*1.5.13. Determine the steady-state temperature distribution between two concen­tric ."phcres with radii 1 and 4, respectively, if thE' temperature of the outersphere is maintained at 80 0 and the inner sphere at. 0° (Sf'f' Exercise 1.5.12).

1.5.14. Isobars are lines of constant temperature. Show that i~obar~ are perpen­dicular to any part of the boundary that is in5ulated.

Orthogonal Curvilinear Coordinates. A coordinate system (u, v, w)may be introduced and defined by x =x(u,v,w),y=y(u,,'.w) and z = z(u,v,w).The radial vector r .::::: xi + Yl + zk. Partial derivatives of r with respect to acoordinate are in the direction of the coordinate. Thus, for example, a vector in theu-direction aT/{) u can be made a unit vector €u in the u-direction by dividing byits length h" = la,./a ul called the scale factor: e" = h'u a,./ au.

*1.5.11. Consider

1.5.12. Assume that the temperature is spherically symmetric 1 v. = u(1'1 t), \vherer is the distance from a fixed point (r2 = .r' + y2 + Z2). Consider the heatflow (without sources) betwf'en any two concentric spheres of radii a and b.

(a) Show that the total heat energy is 471" J~ CpUT2 dr.

(b) Show that the flow of heat energy per unit time out of the spherical shellat r = b is -471"b2 K"a 11/a r I c~b' A similar result holds at r = a.

(c) Use parts (a) and (b) to derive the spherically syrmnetric heat equation

a tL =!':....!!...- ("(2au) .at T'ar ar

1.5.10. Det.E'rmine the equilibrium temperature distribution inside a circle (r :s TO)

if the boundary is fixed at temperature To.

/

/

!I

:JO Chapter 1. Heat Equation 1.5. Heat Equation in Tn-'o or Three Dimensions 31

1&1 = 1. The largest rate of change of 11 (the largest directional derivative) isIv"l > 0, and it occurs if e = 0 (i.e., in the direction of the gradient). Since thisderivative i.s positive, the temperature increase i1'l greatest in the direction of thegradient. Siuce heat energy flO\\'s in the direction of decreasing temperatures, theheat flow vector is in the opposite direction to the heat gradient. It followsthat

(1.5.27)

We have thus learned two properties of the gradient, vu:

1. Direction: 'Vu is perpendicular to the surface u = const.ant. \7u i":l also in thedirection of the largest directional derivative. u increase~ in the direction ofthe gradient.

2. Magnitude: IVI1I is the largest value of the directional derivative.

since l\7ul equals the magnitude of the rate of change of 11 (in the direction of thegradient). This again is callcu Fourier's law of heat conduction. Thus, in threedimensions, the gradient \711 replaces au/Dx.

(1.5.28)

(1.5.29)vl1·(dxi + dyJ + dzk) = o.or

(f)U~ aU. ~ au~) . . ."---J i + -c-j + -3k '(dxi + dyj + dzk) = 0'.r dy z

__-40°

4.5°

.50°

dxi + dyJ + dzk represents any vector in the tangent plane of the level surface.From (1.5.29), its dot product with VI1 is zero; that is, vu is perpendicular to thetangent plane. Thus, 'Vu is perpendicular to the surface u = constant.

Another fundamental property of the gradient is that it is normal (perpendicu­lar) to the level surfaces. It is easier to illustrate this in a two~ dimensional problem(see Fig. 1.5.3) in which the temperature is constant along level curves (rather thanlevel surfaces). To show that the gradient is perpf'ndicular. com,ider the surface allwhich the temperature is the constant To, u(x, Y,Z, t) = T,J. We calculate the dif­ferential of both sides (at a fixed time) along the surface. Since To is constant,dI;l = O. Therefore, using the chain ;rule of partial derivatives,

Figure 1.5.4: The gradient is perpendicular to level ~urfaces of the temperature.

Equation (1.5.28) can be written w

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